Question

以下答案建议使用mysql_但是我决定切换到PDO以提高安全性。但是，当我应用相同的代码时，我很难让它工作。我必须将while切换为foreach以进行PDO。

更新：使用PHP PDO

$sql  = "SELECT * FROM userrecords";
$stmt = $conn->prepare($sql);
$stmt->execute();
$data = $stmt->fetchAll();

 foreach ($data as $row)
 {
  $dateArray = explode('-', $row['eventdate']);
  $year = $dateArray[0];
  $month= $dateArray[1] - 1;
  $day= $dateArray[2];

  $dataArray[] = "[new Date ($year, $month, $day), {$row['scientificname']}, {$row['category_of_taxom']}]";

 }
 echo $dataArray;

Answer 1

基于other question的解决方案，您必须将日期解析为javascript的日期对象格式：

$dataArray = array();
while($row = mysqli_fetch_array($result)) {
    $dateArray = explode('-', $row['Date']);
    $year = $dateArray[0];
    $month = $dateArray[1] - 1; // subtract 1 since javascript months are zero-indexed
    $day = $dateArray[2];
    $dataArray[] = "[new Date($year, $month, $day), {$row['SpeciesA']}, {$row['Speciesb']}]";
}

因为json_encode函数会在这里打破日期，所以我们必须将它解析为DataTable构造函数有点不同：

var data = new google.visualization.DataTable();
data.addColumn('date', 'Date');
data.addColumn('number', 'Species A');
data.addColumn('number', 'Species B');
data.addRows(<?php echo '[' . implode(',', $dataArray) . ']'; ?>);

谷歌图表api日期格式

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